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The emf of hydrogen electrode in term of...

The emf of hydrogen electrode in term of pH is (at 1 atm pressure )

A

`E_(H_2) = (RT //F) xpH `

B

`E_(H_2) = (RT // F) 1// pH`

C

`E_(H_2) = (2.303 RT // F) pH`

D

`E_(H_2) = - 0.0591 pH`

Text Solution

Verified by Experts

The correct Answer is:
D
`E=O-(RT)/(1 x F) ln (1)/((H^(oplus))) -E =(RT)/(F) xx ln (H^(oplus)) - E = (2.303RT)/(F) x - log(H^(oplus)) = P^(H)`
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